Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, x\neq 0$. $\dfrac{{(r)^{-4}}}{{(r^{3}x^{-4})^{-3}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r}$ to the exponent ${-4}$ . Now ${1 \times -4 = -4}$ , so ${(r)^{-4} = r^{-4}}$ In the denominator, we can use the distributive property of exponents. ${(r^{3}x^{-4})^{-3} = (r^{3})^{-3}(x^{-4})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r)^{-4}}}{{(r^{3}x^{-4})^{-3}}} = \dfrac{{r^{-4}}}{{r^{-9}x^{12}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-4}}}{{r^{-9}x^{12}}} = \dfrac{{r^{-4}}}{{r^{-9}}} \cdot \dfrac{{1}}{{x^{12}}} = r^{{-4} - {(-9)}} \cdot x^{- {12}} = r^{5}x^{-12}$.